Doug McIlroy <d...@cs.dartmouth.edu> wrote:
> instance Num a => Num (a,a) where
> (x,y) * (u,v) = (x*u-y*v, x*v+y*u)
>
> Unfortunately, type inference isn't strong enough to cope with
>
> (1,1)*(1,1)
I'm guessing it is because
(fromInteger 1, fromInteger 1) :: (Num a, Num b) => (a,b)
So you want to force the two components to be the same type:
let x = 1 in (x,x)*(x,x)
Or for differing component values:
let x = 1 in (x,2`asTypeOf`x)*(3`asTypeOf`x,4`asTypeOf`x)
Regards,
Malcolm
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